In hyperbolic geometry, we know that any complete hyperbolic surface is the quotient $\mathbb{H}^{2}/\Gamma$ where $\Gamma < \text{Isom}(\mathbb{H}^{2})$ a discrete subgroup.

We know $\text{Isom}^{+}(\mathbb{H}) \cong PSL(2,\mathbb{R})$ which contains $PSL(2,\mathbb{Z})$ as a discrete subgroup. Also $\Gamma(m)$ which is the kernel of the natual map $PSL(2,\mathbb{Z}) \to SL(2,\mathbb{Z}/m\mathbb{Z})$ acts freely and properly discontinuously on $\mathbb{H}^{2}$ so $\mathbb{H}^{2}/\Gamma(m)$ is a hyperbolic surface. I want to see a geometric picture of this manifold. How can I proceed to see what this manifold look like (at least topologically, if not isometrically)?

1 Answer
1

This is related to very deep questions in number theory, and there's no simple answers.

From a naive point of view it seems reasonable. You can start with a fundamental domain for the fractional linear action of $PSL(2,\mathbb Z)$ on the upper half plane; the conventional fundamental domain is given by intersecting $-\frac{1}{2} \le \text{Real}(z) < \frac{1}{2}$ with $\text{abs}{z} \ge 1$. Then you can carefully choose coset representatives of $\Gamma(m)$ in $PSL(2,\mathbb Z)$, use them to translate the fundamental domain around, and hope that you chose those representatives carefully enough so that the union of those translated fundamental domains is a polygon which you can then glue up to get $\mathbb H^2 / \Gamma(m)$. For the case $m=2$ this can probably be done by hand. But the index of $\Gamma(m)$ in $PSL(2,\mathbb Z)$ grows uncomfortably fast: it is cubic in $m$. So this method is not practical in general.

The general heading of your question is the topic of congruence subgroups of $PSL(2,\mathbb Z)$. In that reference you'll see, for example, a link to a 2006 paper by Long, Machlachlan and Reid proving that there are only finitely many values of $m$ such that the quotient surface $\mathbb H^2 / \Gamma(m)$ has genus zero, meaning that it is homeomorphic to a sphere with some finite number of points removed. You'll also see described some amazing connections between the topology of $\mathbb H^2 / \Gamma(m)$ (more properly, not that surface itself but a closely related quotient orbifold) and the prime factors of the order of the monster group!

$\begingroup$I was not expecting that surfaces that can be described so easily could be so monstrous. But is there really no way to tell anything about the surface $\mathbb{H}^{2}/\Gamma(m)$?$\endgroup$
– Prakhar GuptaDec 1 '18 at 17:04

3

$\begingroup$Well, that's what the second paragraph of my answer does: it tells something about the surfaces $\mathbb H^2 / \Gamma(m)$. And there's still more to learn about them from the theory of congruence subgroups. But you have to work hard to understand the statements and their proofs. It's true: sometimes mathematical objects that are relatively simple to define are very difficult to say something substantial about.$\endgroup$
– Lee MosherDec 3 '18 at 22:01